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A319594
Number of solutions to dft(p)^2 + dft(q)^2 = (4n-3), where p and q are even sequences of length 2n-1, p(0)=0, p(k)=+1,-1 when k<>0, q(k) is +1,-1 for all k, and dft(x) denotes the discrete Fourier transform of x.
5
2, 4, 4, 12, 12, 0, 12, 16, 0, 36, 24, 0, 20, 36, 0, 60
OFFSET
1,1
COMMENTS
Each solution (p,q) corresponds to a family of symmetric Hadamard matrices of size 8n-4. To construct one member from this family, set A = circulant(p) + I, B = circulant(q), C = B, D = A - 2 I and H = [ [A, B, C, D], [B, D, -A, -C], [C, -A, -D, B], [D, -C, B, -A]]. Then A, B, C and D are symmetric and H is Hadamard and symmetric.
Since p and q are assumed to be even, dft(p) and dft(q) are real-valued.
2 divides a(n) for all n. If (p,q) is a solution, then (p,-q) is also a solution.
4 divides a(n) when n>1. If (p,q) is a solution, then (+/-p,+/-q) are also solutions. When n=1, p is the length-1 sequence, (0).
It is known that a(n)>0 for n=25, 26, 29.
LINKS
Jeffery Kline, List of all pairs (p,q) that are counted by a(n), for 1<=n<=16.
Jeffery Kline, List of pairs (p,q) that establish a(n)>0, for n=25, 26, and 29.
Jeffery Kline, Geometric Search for Hadamard Matrices, Theoret. Comput. Sci. 778 (2019), 33-46.
J. Seberry and N.A. Balonin, The Propus Construction for Symmetric Hadamard Matrices, arXiv:1512.01732 [math.CO], 2015.
J. Seberry and N.A. Balonin, Two infinite families of symmetric Hadamard matrices, Australasian Journal of Combinatorics, 69 (2015), 349-357.
EXAMPLE
For n=1, the a(1)=2 solutions are ((0),(-)) and ((0),(+)). For n=2, the a(2)=4 solutions are ((0,-,-), (-,+,+)), ((0,-,-), (+,-,-)), ((0,+,+),(-,+,+)) and ((0,+,+), (+,-,-)).
KEYWORD
nonn,more
AUTHOR
Jeffery Kline, Dec 16 2018
STATUS
approved